poj3518 Prime Gap

最新推荐文章于 2021-10-11 16:57:25 发布

不会敲代码的小帅

最新推荐文章于 2021-10-11 16:57:25 发布

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本文链接：https://blog.youkuaiyun.com/SunPeishuai/article/details/81586519

本文介绍了一个计算特定整数所在素数间隙长度的程序设计方法。该程序能够接收一个正整数作为输入，并输出该整数所在素数间隙的长度，如果输入的数是素数，则输出0。

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Prime Gap

Description

The sequence of n − 1 consecutive composite numbers (positive integers that are not prime and not equal to 1) lying between two successive prime numbers p and p + n is called a prime gap of length n. For example, ‹24, 25, 26, 27, 28› between 23 and 29 is a prime gap of length 6.

Your mission is to write a program to calculate, for a given positive integer k, the length of the prime gap that contains k. For convenience, the length is considered 0 in case no prime gap contains k.

Input

The input is a sequence of lines each of which contains a single positive integer. Each positive integer is greater than 1 and less than or equal to the 100000th prime number, which is 1299709. The end of the input is indicated by a line containing a single zero.

Output

The output should be composed of lines each of which contains a single non-negative integer. It is the length of the prime gap that contains the corresponding positive integer in the input if it is a composite number, or 0 otherwise. No other characters should occur in the output.

Sample Input

Sample Output

题意：输入一个数，若为素数则输出0，非素数时，用向后到达的第一素数减去向前的第一个素数

思路：先把素数进行打表标记，通过两个for循环分别向前向后查找，输出相减的结果。

代码：

#include<cstdio>
#include<iostream>
using namespace std;
const int maxn= 1399710;
int isprime[maxn];
void prime()//是素数的都为0
{
    for(int i=2;i<=maxn;i++)
    {
        if(!(isprime[i]))
        {
            isprime[i]=0;
        }
        for(int j=i*2;j<=maxn;j+=i)
        {
            isprime[j]=1;
        }
    }
}
int main()
{
    int  n;
    prime();
    while(scanf("%d",&n)!=EOF)
    {
        if(n==0)
            break;
        if(isprime[n]==0)
        {
            printf("0\n");
            continue;
        }
        else
        {
            int left,right;
         for(left=n;isprime[left];left--);
         for(right=n;isprime[right];right++);
         printf("%d\n",right-left);
        }
    }
    return 0;
}